Problem: The value of $\sqrt{128}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Answer: Consider the perfect squares near $128$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $121$ is the nearest perfect square less than $128$ $144$ is the nearest perfect square more than $128$ So, we know $121 < 128 < 144$ So, $\sqrt{121} < \sqrt{128} < \sqrt{144}$ So $\sqrt{128}$ is between $11$ and $12$.